June 2018: Paper 3, Unified chemistry

Mark scheme

Examiner's comment

Most candidates were credited this straightforward mark and identified that hydrogen would gain an electron to form a 1– ion. Some candidates opted for lithium, able to form an ion with the same electron configuration as helium, but with a 1+ rather than a 1– charge. Candidates are recommended to look closely at the requirements of the question set.

Notes from the mark scheme and report

• hydrogen (H– is 1s2)
• ALLOW H2

Mark scheme

Examiner's comment

This part required candidates to recall their knowledge of trends in first ionisation energy. Candidates found this part harder than 1(a)(i) with only the higher ability candidates choosing the correct response of ‘helium’. Many candidates instead chose another noble gas, with neon and argon commonly seen. Other common incorrect responses were hydrogen and fluorine.

Notes from the mark scheme and report

• helium

Mark scheme

Examiner's comment

Most candidates did correctly select magnesium, but many other elements were seen, especially aluminium, silicon, beryllium and calcium. To identify the element’s group, candidates needed to analyse the data to find the large increase in ionisation energy corresponding to a change in shell. From the responses, some candidates did not make use of ‘Period 3’ in the stem.

Notes from the mark scheme and report

• magnesium - large jump after second ionisation energy

Skills: M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Examiner's comment

Most candidates selected sulfur as the correct response, recalling their knowledge of molecular shapes encountered early in the course. There was no real pattern for incorrect responses, suggesting that they were guesses.

Notes from the mark scheme and report

• sulfur (SF6)
• ALLOW S8

Skills: M4.1 Use angles and shapes in regular 2-D and 3-D structures

Mark scheme

Examiner's comment

Most candidates chose the correct response of chlorine, although hydrogen was a common incorrect response, presumably by linking to the acidic properties of H+ ions. Other candidates focused on ‘reacts with water’ and chose sodium (which does form a solution with water, but on that is alkaline rather than acidic).

Notes from the mark scheme and report

• chlorine OR fluorine

Mark scheme

Examiner's comment

Almost all candidates correctly responded with phosphorus and this was the easiest part of 1(a).

Notes from the mark scheme and report

• phosphorus (31.0 + 3.0)
• ALLOW P4

Skills: M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

Most candidates correctly selected carbon. From their A Level studies, candidates would expect hydrogen to have an oxidation number of +1 and to form compounds with carbon (CH4) and silicon (SiH4) in which the element has an oxidation number of –4. Although hydrogen is actually slightly less electronegative than carbon, hydrogen is slightly more electronegative than silicon. Therefore, in the case of SiH4, silicon has an oxidation number of +4. A response of silicon still indicates a correct understanding of oxidation number rules and was also credited

Notes from the mark scheme and report

• carbon (CH4)
• ALLOW silicon

Mark scheme

Examiner's comment

This proved to be the hardest part of 1(a) with only the higher ability candidates selecting oxygen. Sulfur proved to be the key distractor, having the same molar mass as O2. Most candidates did not consider that the element was gaseous and could not be sulfur.

Notes from the mark scheme and report

• oxygen: 1.33 × 10^-3 × 24 000 = 32 g mol–1 so O2

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.2 Change the subject of an equation

Mark scheme

Examiner's comment

This part required candidates to determine the structure and bonding of unfamiliar compounds from their melting points and electrical conductivities. This part discriminated very well. Candidates with a good knowledge and understanding of structure and bonding, often produced concise, clear responses which were credited with a high mark, as shown in the Poor understanding often showed up with contradictions and it was common to see ionic lattices with strong intermolecular forces and able to release mobile electrons (rather than mobile ions) in the liquid state. Weaker candidates often wrote extensive answers, although the extra length did not lead to more marks as correct responses were then often contradicted. Exemplar 2 shows part of a very muddled and rambling response in which NaCl is described as having both ionic bonding and London forces. The best responses identified NaCl and MgCl2 as having giant ionic lattices with strong forces between their ions. This structure was then compared with the simple molecular structures of SiCl4, PCl3 and SCl2 with weak London forces between their molecules. A common error was for SiCl4 to have a giant covalent lattice, presumably linking with the structure of elemental silicon and not using the low boiling point in the supplied data.

Notes from the mark scheme and report

• NaCl/MgCl2 giant ionic with ions mobile in LIQUID (IGNORE 'free ions' and aqueous)
• SiCl4/PCl3/SCl2 simple molecular with London forces (IGNORE permanent dipole and van der Waals)
• ionic bonds stronger than London forces

Mark scheme

Examiner's comment

The graph was generally well drawn although many didn’t realise that a tangent needed to be drawn. A few tried to draw a straight line of best fit, rather than a curve. Most candidates were able to construct a correct graph, assigning the axes correctly and making good use of the graph paper. Points were usually plotted correctly and most candidates made a good attempt to draw a curved line of best fit. Candidates then needed to draw a tangent at 500 seconds and to measure its gradient to calculate the rate. Lower ability candidates did not draw a tangent to their curve and just calculated the volume formed at 500 seconds.

Notes from the mark scheme and report

• axes labelled with units, linear scales using at least half the grid
• smooth curve through origin not point-to-point lines
• tangent at t = 500 ± 100 s
• rate 0.037–0.047
• using x and y coordinates at 500 s (33/500 = 0.066) scores zero

Skills: M3.1 Translate information between graphical, numerical and algebraic forms; M3.2 Plot two variables from experimental or other data; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change; PAG9 Rates of reaction – continuous monitoring

Mark scheme

Examiner's comment

Most candidates calculated the initial concentration of 0.092 mol dm–3, although many ignored the two significant figures requirement with 0.0916 and 0.09 being seen commonly. Almost all candidates corrected calculated that 55 cm3 O2 contains 2.29 × 10-3 mol O2(g) and multiplied this value by 2 to obtain 4.58 × 10–3 mol H2O2 in 50 cm3 of solution. Fewer candidates scaled up this value to 1 dm3 with many dividing by 20 (instead of × 20) or using 55 cm3, the initial volume of O2(g) rather than the volume of the H2O2 solution. possible though to see that the scaling up by 2 for H2O2 has been omitted and the final answer (0.046) is half of the correct answer (0.092). It is then possible to award the final mark by error carried forward.

Notes from the mark scheme and report

• common error 0.046 (no ×2 for n(H2O2)) scores 2
• 2 SF required (0.092)
• ALLOW ideal gas equation at 293–298 K

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This part discriminated extremely well with many candidates obtaining either both marks or zero marks. Candidates needed to select the correct redox pairs, combine them and cancel H+ and H2O. Many otherwise correct equations did not receive credit as they contained basic balancing errors. As a final step in constructing an equation, candidates are recommended to check the formulae of their species and that the equation is balanced.

Notes from the mark scheme and report

• 2MnO4– + 5H2O2 + 6H+ → 2Mn2+ + 8H2O + 5O2
• first mark for balanced but uncancelled H+/e–
• 1 mark only for 2H2O2 → 2H2O + O2

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

In this part, candidates needed to apply their knowledge and understanding of ligand substitution and equilibrium to a novel situation. The best equations used Cl– ions to form CoCl4 2–. Some candidates used HCl instead and then H+ was often omitted in the equation. As with 2b, candidates are recommended to check that their completed equations are balanced.

Notes from the mark scheme and report

• ALLOW reverse equation
• IGNORE state symbols

Mark scheme

Examiner's comment

Candidates were expected to determine the type of energy change by linking their equilibrium equation in 2b(i) with the colour changes at different temperatures. Most candidates correctly concluded that the formation of a blue colour is endothermic. Many candidates did not explain this in terms of a shift in equilibrium position, considering bond breaking and bond making instead.

Notes from the mark scheme and report

• mark independently
• equilibrium shifts right at high temperature (temperature must be stated) or left at 0 °C - direction depends on equation written in (c)(i)
• endothermic
• IGNORE 'shifts to blue side'

Mark scheme

Examiner's comment

Most candidates were able to draw clear octahedral 3D shapes of the optical isomers, with good use made of wedges. The best candidates showed an overall 3– charge on each isomer but lower ability candidates did not always cancel down the charges in the Fe3+ and (COO–)2 ions, showing a 3+ charge instead of the overall charge of 3–. (See Exemplar 5)

Notes from the mark scheme and report

• overall 3– charge shown outside brackets
• one mark per isomer
• bonds must go to O atoms of each ligand
• 3D wedges required
• simplified loops for ligands NOT allowed

Skills: M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects; M4.3 Understand the symmetry of 2-D and 3-D shapes

Mark scheme

Examiner's comment

Candidates were expected to apply their knowledge of the colour change in a manganate(VII) titration to this novel situation. Most candidates incorrectly showed the inverse colour change of yellow → colourless.

Notes from the mark scheme and report

• colourless to yellow (first excess Ce4+)
• IGNORE clear

Mark scheme

Examiner's comment

Most candidates answered this unstructured titration calculation well, with almost all responses gaining some credit. Most coped well with the 1:2 reaction stoichiometry and the need to scale up to 250 cm3. Common errors included the following. • Taking the mean of all three titres (23.30 cm3) instead of the closest titres (23.20 cm3). • Not considering the 1:2 stoichiometry. • Not scaling up to 250 cm3. • Giving the final answer to two rather than three significant figures; this was the most common error. The question required the final answer to be given to an appropriate number of significant figures. Many candidates seemed to be unaware that this reflects the least significant figures provided in the data, in this case three significant figures. Candidates are also advised to only round at the end of a multi-step calculation. Rounding of intermediate values introduces rounding errors in the final answer. The example shows a perfect response with each step clearly described, leading to the correct concentration of ethanedioic acid. Candidates should present their calculations clearly. If there is a mistake at any stage, this can be easily identified, allowing for error carried forward to be applied for any subsequent good method.

Notes from the mark scheme and report

• common errors: mean of all three titres 23.30 gives 0.634% (5 marks)
• no ÷2 gives 1.26% (5 marks)
• 0.63 loses final mark (must be 3 SF)
• mean from concordant titres 23.15 and 23.25 only
• ×10 scale-up

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M1.2 Find arithmetic means; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Almost all candidates responded with the correct oxidation number of +2. Incorrect responses were 0 (the overall charge of the complex), +4 or 2 (with no sign). Candidates are reminded on the importance of the sign in assigning oxidation numbers.

Notes from the mark scheme and report

• +2
• sign required

Mark scheme

Examiner's comment

This part required candidates to apply their knowledge and understanding of the nucleophilic substitution mechanism in a novel context. This part discriminated extremely well with able candidates drawing the correct initial complex and showing precisely positioned curly arrows, dipoles and the role of the NH3 lone pair, as shown in Lower ability candidates often showed imprecise curly arrows or placed a negative charge on NH3, as shown in Exemplar 8.

Notes from the mark scheme and report

• curly arrow from lone pair on N of NH3 to Pt
• DO NOT ALLOW charge on NH3
• ALLOW SN1-type route with + on Pt intermediate
• IGNORE dipoles

Mark scheme

Examiner's comment

This part assessed knowledge of functional groups and proved to be a very good discriminator. Able candidates usually identified the phenol and amide functional groups, with ‘secondary amide’ also seen. In Exemplar 9, the candidate has identified the correct functional groups. The candidate’s working by circling the functional groups in the structure shows good examination technique, helping the candidate to arrive at the correct conclusion. The phenol group was often incorrectly identified as an alcohol and the amide group as a combination of ‘amine’, ‘ketone’, ‘keytone’ or ‘carbonyl’. Neutral responses such as ‘hydroxyl’ and ‘benzene’ were ignored. Candidates need to be careful that they do not present an extensive list of many functional groups in the hope that the correct groups are amongst them, as shown in Exemplar 10. Incorrect groups are marked first.

Notes from the mark scheme and report

• phenol and amide
• if more than 2 groups given mark incorrect ones first
• IGNORE hydroxyl/hydroxy, aryl, benzene
• ketone/carbonyl is incorrect

Mark scheme, with the levels and indicative points

Examiner's comment

This part assessed practical aspects of a two-stage organic synthesis. Overall, candidates responded well, and this part was discriminating. Many candidates produced well-structured responses although lower ability candidates do have problems with constructing a cohesive response. Most candidates identified the correct reagents (Sn and concentrated HCl) and the intermediate (4- aminophenol), which was usually shown as its structure. Able candidates usually showed that 11.5 g of 4-nitrophenol is needed for the synthesis. A common error used the ‘inverse percentage’ ratio of 40/100, resulting in an incorrect mass of 1.84 g. Candidates are recommended to check whether a calculated answer looks sensible. Looking at the structures and with a percentage yield of 40%, 1.84 g does not look to be enough of the starting chemical. Some lower ability candidate responses assumed that 5.00 g was 40% of the required mass and responded with 5.00 × 100/40 = 12.5 g. There were some good descriptions of purification, although finer details such as using a minimum volume of hot solvent, washing with cold solvent, and drying) were often omitted. Candidates needed to respond with more than just ‘recrystallisation’. In the purification, common errors were showing the correct steps but in the wrong order and use of a drying agent such as CaCl2 (confusion with part of the purification of an organic liquid). These candidates seemed unaware that adding a solid drying agent to an organic solid would result in impure paracetamol rather than purifying. In comparison, Exemplar 12 is much less detailed: concentrated HCl has not been shown as a reagent for step 1, the candidate has not shown that they know how to carry out a percentage yield calculation, and the purification is confused, and lacks detail.

Notes from the mark scheme and report

• 5.00/151 = 0.0331 mol, ×100/40 = 0.0828 mol, × 139 = 11.5 g (inverted ratio gives 1.84 g)
• Sn + conc HCl then NaOH
• intermediate 4-aminophenol
• recrystallise: dissolve in MINIMUM volume of HOT solvent, cool, filter, wash with cold solvent, dry
• 'recrystallisation' alone is not a detailed description

Skills: M0.2 Use ratios, fractions and percentages; M2.2 Change the subject of an equation; PAG6 Synthesis of an organic solid

Mark scheme

Examiner's comment

The direct determination of an enthalpy change is a common practical procedure. Most candidates were able to calculate the energy change in the reaction as 2.09 kJ, with comparatively few not adding the two 50 cm3 volumes in their calculation. Common errors were using the mass or moles of succinic acid in the mc∆T calculation or to use 278 K, rather than 5ºC for ∆T. Most candidates calculated that 0.02 mol of succinic acid had reacted. The best responses recognised that the enthalpy change of neutralisation (i.e. formation of 1 mol of H2O) was required and that 0.02 mol succinic acid produced 2 × 0.02 = 0.04 mol H2O. 2.09 kJ was then scaled up to give the correct ∆H value of –52.25 kJ mol–1. Example 1 follows this approach and each stage in the working can be followed easily. Most candidates scaled up using 0.02 mol to obtain –104.6 kJ mol–1 and then thought that this was the final answer. Lower ability candidates often calculated the moles of succinic acid as 0.02 but also calculated the moles of NaOH as 0.05 mol. They often then went on to scale up their energy change by 0.05 or 0.07. Many responses were disorganised and poorly presented with unsubstantiated numbers across the answer space. Imperfect calculations may be partially credited by applying error carried forward but this is only possible if sense can be made of the response. Compare the responses in Exemplar 13 and Exemplar 14. In Exemplar 14, there is no labelling to show what each stage refers to. It looks as if the candidate has calculated the moles of NaOH rather than succinic acid and has used this in the subsequent stage. Error carried forward has been applied generously on this assumption. Answer = –52.25 kJ mol–1.

Notes from the mark scheme and report

• –52.3 from two cancelling errors (50 g instead of 100 g AND no ÷2) scores only 2
• succinic acid limiting, 2 mol H2O per mol acid
• DO NOT ALLOW fewer than 3 SF
• – sign required

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

Candidates found this part difficult and only higher ability candidates identified that a titration could easily determine the concentration of succinic acid. The answers seen covered most of the techniques encountered in the course. Candidates should consider the information provided in a practical context to arrive at an informed response rather than what sometimes seemed to be a guess.

Notes from the mark scheme and report

• titration
• IGNORE type of titration

Skills: PAG2 Acid–base titration

Mark scheme

Examiner's comment

Candidates were required to derive the equation from which the supplied Kc expression had been written. Overall, this part was answered well but some candidates struggled with the brackets or used CH2COOH2 for succinic acid.

Notes from the mark scheme and report

• (CH2COOH)2 + 2C2H5OH ⇌ (CH2COOC2H5)2 + 2H2O
• ALLOW → and molecular formulae

Mark scheme

Examiner's comment

This part discriminated extremely well with many candidates finding it difficult to convert the bracketed structural formula into a skeletal formula. Common errors were drawing of the mono-ester or omitting a carbon atom in the centre of the structure. Even when incorrect, most attempted answers were skeletal formulae.

Notes from the mark scheme and report

• skeletal diethyl butanedioate
• IGNORE displayed formulae

Mark scheme

Examiner's comment

Many candidates did not seem to realise that the supplied equation used moles, not concentrations. Those who did often stated that the mole representation could be used because the volume was the same for all. Of those who went on to state that the volume would cancel, only a few explained why that was true in this particular case. This challenging part discriminated very well. The best responses showed the units as n/V in the expression and showed that the volumes cancel. Question 5b(v) Overall, this part discriminated well with many candidates obtaining the correct answer of 0.67. Common errors included a one significant figure answer of 0.6 or 0.7 and 0.375, by using 0.12 mol instead of 0.09 mol for the moles of ethanol. Many successful answers were well-presented and included a table of initial and final values. This gave a systematic way of deriving the equilibrium moles.

Notes from the mark scheme and report

• volume cancels / same number of moles on each side of the equation
• IGNORE 'volume is the same' and 'Kc has no units'

Mark scheme

Examiner's comment

A LEVEL H432/03 Summer 2018 series Version 1 H432 For first teaching in 2015 CHEMISTRY A Examiners’ report www.ocr.org.uk/chemistry Contents Introduction .............................................................................................................................................. 3 Paper H432/03 series overview ................................................................................................................ 4 Question 1(a)(i) ..................................................................................................................................... 5 Question 1(a)(ii) .................................................................................................................................... 5 Question 1(a)(iii) ................................................................................................................................... 5 Question 1(a)(iv) ................................................................................................................................... 6 Question 1(a)(v) .................................................................................................................................... 6 Question 1(a)(vi) ................................................................................................................................... 6 Question 1(a)(vii) .................................................................................................................................. 6 Question 1(a)(viii).................................................................................................................................. 7 Question 1(b) ........................................................................................................................................ 7 Question 2(a)(i) ................................................................................................................................... 10 Question 2(a)(ii) .................................................................................................................................. 11 Question 2(b) ...................................................................................................................................... 13 Question 2(c)(i) ................................................................................................................................... 14 Question 2(c)(ii) .................................................................................................................................. 14 Question 3(a) ...................................................................................................................................... 15 Question 3(b)(i) ................................................................................................................................... 16 Question 3(b)(ii) .................................................................................................................................. 17 Question 4(a)(i) ................................................................................................................................... 18 Question 4(a)(ii) .................................................................................................................................. 19 Question 4(b)(i) ................................................................................................................................... 20 Question 4(b)(ii)* ................................................................................................................................. 21 Question 5(a) ...................................................................................................................................... 25 Question 5(b)(i) ................................................................................................................................... 27 Question 5(b)(ii) .................................................................................................................................. 27 Question 5(b)(iii) ................................................................................................................................. 27 Question 5(b)(iv) ................................................................................................................................. 28 Question 5b(v) .................................................................................................................................... 28 Question 6(a)(i) ................................................................................................................................... 29 Question 6(a)(ii) .................................................................................................................................. 29 Question 6(a)(iii) ................................................................................................................................. 30 Question 6(a)(iv) ................................................................................................................................. 31 Question 6(b)* ..................................................................................................................................... 31 Introduction Our examiners’ reports are produced to offer constructive feedback on candidates’ performance in the examinations. Question 5b(v) Overall, this part discriminated well with many candidates obtaining the correct answer of 0.67.

Notes from the mark scheme and report

• equilibrium moles: ester 0.0300, water 0.0600, ethanol 0.150 – 0.060 = 0.0900
• Kc = 0.667
• 0.02 must be used for acid
• ALLOW ECF
• IGNORE units

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Most candidates made good attempts at the name, the difficulty being that hydroxyl group needed to be shown as a hydroxy- prefix, rather than the suffix -ol. Common errors included 2-hydroxybutanal (counting the carbon chain from the wrong end) and 2- or 3-hydroxybutanoic acid (reading the aldehyde group as a carboxylic acid).

Notes from the mark scheme and report

• 3-hydroxybutanal
• DO NOT ALLOW 3-hydroxybutal or 3-hydroxylbutanal
• IGNORE hyphens/commas

Mark scheme

Examiner's comment

This part was answered well with most choosing nucleophilic addition. Credit was given just for ‘addition’.

Notes from the mark scheme and report

• addition
• IGNORE nucleophilic/electrophilic
• DO NOT ALLOW addition–elimination, condensation or polymerisation

Mark scheme

Examiner's comment

This novel question linked together acid–base equilibria with a multi-step process. Many candidates completed an equation to generate acid–base pairs, which were then usually assigned correctly. The final equation was challenging but the highest ability candidates were able to combine together all the information with their earlier responses to arrive at the correct equation. See Exemplar 15.

Notes from the mark scheme and report

• step 1 CH3CHO + OH– ⇌ –CH2CHO + H2O with pairs A1/B1 and A2/B2 labelled
• step 2 CH3CHO + –CH2CHO + H2O → CH3CHOHCH2CHO + OH– (catalyst regenerated)
• ALLOW H+ transfer from OH– version
• IGNORE connectivity

Mark scheme

Examiner's comment

This part was one of the most challenging on the paper. Candidates needed to link the earlier information for combining two ethanal molecules to derive the product for combining two propanone molecules. Despite the challenge, the highest ability candidates were able to come up with the correct structure.

Notes from the mark scheme and report

• (CH3)2C(OH)CH2COCH3
• connectivity not assessed

Mark scheme, with the levels and indicative points

Examiner's comment

Overall, this part was answered well. Good answers were well organised, showing clear mechanisms with precisely positioned curly arrows. Most candidates included a clear definition of an electrophile and were able to select appropriate reactions. Most candidates were familiar with the mechanisms for electrophilic addition and electrophilic substitution. Mechanisms of an alkene with HBr or Br2 and benzene with NO2 + or Br+ were the most commonly seen. Most candidates showed the role of a catalyst in electrophilic substitution. Common errors/omissions were the direction of the curly arrow from the aromatic ring to the electrophile, the position of the curly arrow when H+ is lost from an aromatic intermediate, and not showing the lone pair when Br- attacks a carbocation. Some answers lacked detail and gave only two mechanisms with minimal supporting words. Lower ability candidates described nucleophilic addition or substitution as one of their mechanisms or had curly arrows going in the wrong direction. A few candidates answered in prose without including equations or diagrams. Candidates are advised that mechanisms must always be communicated in the usual diagrammatic way. and electrophilic substitution, including the role of the FeCl3 catalyst. Curly arrows are precisely positioned, with correct use of lone pairs and charges. The candidate has demonstrated excellent knowledge and understanding. The response in Exemplar 17 is clearly at a different level. The candidate has chosen an alkane rather than an alkene and has used curly arrows and charges incorrectly. This candidate appears to have been poorly prepared.

Notes from the mark scheme and report

• electrophile as electron pair acceptor
• aliphatic e.g. ethene + Br2 electrophilic addition with dipole, curly arrows and carbocation
• aromatic e.g. nitration/halogenation of benzene electrophilic substitution with generation of electrophile and regeneration of catalyst
• accurately positioned curly arrows and charges